**Mochizuki's Frobenioid reconstruction: the final bit**In "The geometry of Frobenioids 1"

http://goo.gl/3tWTC Mochizuki 'dismantles' arithmetic schemes and replaces them by huge categories called Frobenioids. Clearly, one then wants to reconstruct the schemes from these categories and we almost understood how he manages to to this ( see

http://goo.gl/TbMLl) modulo the 'problem' that there might be auto-equivalences of Frob(Z), the Frobenioid corresponding to Spec(Z) i.e.the collection of all prime numbers, reshuffling distinct prime numbers.

In previous posts i've simplified things a lot, leaving out the 'Arakelov' information contained in the infinite primes, and feared that this lost info might be crucial to understand the final bit. Mochizuki's email

http://goo.gl/42dd3 also points in that direction.

Here's what i hope to have learned this week:

**the full Frobenioid Frob(Z)****objects** of Frob(Z) consist of pairs (q,r) where q is a strictly positive rational number and r is a real number.

**morphisms** are of the form f=(n,a,(z,s)) : (q,r) --> (q',r') (where n and z are strictly positive integers, a a strictly positive rational number and s a positive real number) subject to the relations that

q^n.z = q'.a and n.r+s = r'+log(a)

n is called the Frobenious degree of f and (z,s) the divisor of f.

All this may look horribly complicated until you realise that the isomorphism classes in Frob(Z) are exactly the 'curves' C(alpha) consisting of all pairs (q,r) such that r-log(q)=alpha, and that morphisms with the same parameters as f send points in C(alpha) to points in C(beta) where

beta = n.alpha + s - log(a)

So, the isomorphism classes can be identified with the real numbers R and special linear morphisms of type (1,a,(1,s)) and their compositions are compatible with the order-structure and addition on R.

Crucial is Mochizuki's observation that C(0) are precisely the 'Frobenious-trivial' objects in Frob(Z) (i'll spare you the details but it is a property on having sufficiently many nice endomorphisms).

Now, consider an auto-equivalence E of Frob(Z). It will induce a map between the isoclasses E : R --> R which is additive and as Frob-trivs are mapped under E to Frob-trivs this will map 0 to 0, so E will be an additive group-endomorphism on R hence of the form x --> r.x for some fixed real number r, and we want to show that r=1.

Linear irreducible morphisms are of type (1,a,(p,0)) where p is a prime number and they map C(alpha) to C(alpha-log(p)). As irreducibles are preserved under equivalence this means that for each prime p there must exist a prime q such that r log(p) = log(q)

Now if r is an irrational number, there must be at least three triples (p1,q1),(p2,q2) and (p3,q3) satisfying r log(pi) = log(qi) but this contradicts a fairly hard result due to Lang that for 6 distinct primes l1,...,l6 there do not exist positive rational numbers a,b such that

log(l1)/log(l2) = a log(l3)/log(l4) = b log(l5)/log(l6)

So, r must be rational and of the form n/m, but then r=1 (if not the correspondence r.log(p) = log(q) gives p^n=q^m contradicting unique factorisation). This then shows that under the auto-equivalence each prime p (corresponding to a linear irreducible map) is send to itself.

This was the remaining bit left to show (as in

http://goo.gl/TbMLl) that the Frobenioid corresponding to any Galois extension of the rationals contains enough information to reconstruct from it the schemes of all rings of integers in intermediate fields.